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I know this is a simple question but i'm not sure of which combinatorial selection equation to use.
How many distinct binary bit strings of length fifteen are there?
Using a simple example, would someone be able to explain the difference between
I have had a look around but none of the examples are making much sense.
With binary strings of length $15$ you can't have without repetition. There are only two characters, so by the third character you will have repetition. If the string is not ordered you only care how many $0$s and $1$s there are. The number of $0$s can range from $0$ to $15$-how many is that? Then the number of $1$s is determined. For ordered strings, each bit has two choices, so there are $2^{15}$
With repetition: you are allowed to choose the same item more than once. For example, $ABC$ is an allowable choice of three items, and so is $AAC$.
Without repetition: you are not allowed to choose the same item more than once. For example, $ABC$ is still an allowable choice of three items, but $AAC$ is not.
Ordered: the order of items makes a difference. For example, $ABC$ and $CBA$ are different selections and are both counted. Likewise $AAC$ and $ACA$.
Unordered: the order of items makes no difference. For example, $ABC$, $ACB$, $BAC$, $BCA$, $CAB$, $CBA$ are all regarded as the same and therefore count as one possibility, not six. Likewise, $AAC$, $ACA$, $CAA$ count as one possibility, not three.
If you understand what a bit string is, I'm sure you can now work out whether repetition is allowed or not, and whether order is important or not.
